A cylindrical tank with radius $7 \mathrm{~m}$ is being filled with water at a rate of $4 \mathrm{~m}^{3} / \mathrm{min}$. How fast is the height of the water increasing (in $\mathrm{m} / \mathrm{min}$ )? $\frac{4}{9 \pi}$ $\mathrm{m} / \mathrm{min}$

Step 1 ：Given the rate of volume increase $\frac{dV}{dt} = 4 \mathrm{~m}^3/\mathrm{min}$ and the radius of the cylindrical tank $r = 7 \mathrm{~m}$, we need to find the rate of height increase $\frac{dh}{dt}$. The volume of a cylinder is given by $V = \pi r^2 h$. Differentiating both sides with respect to time, we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.

Step 2 ：Plugging in the given values, we have $4 = \pi (7^2) \frac{dh}{dt}$. Solving for $\frac{dh}{dt}$, we get $\boxed{\frac{dh}{dt} = \frac{4}{49 \pi} \mathrm{m} / \mathrm{min}}$.